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POIs there a simpler way than this to calculate a straight in poker?
 

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  1. POIs there a simpler way than this to calculate a straight in poker?
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    <p>I have an algorithm for calculating whether a player's hand holds a straight in Texas Hold'em. It works fine, but I wonder if there is a simpler way to do it that does not involve array/string conversions, etc.</p> <p>Here's a simplified version of what I have. Say the player is dealt a hand that is a 52-element array of card values:</p> <pre><code>var rawHand = [1,0,0,0,0,0,0,0,0,0,0,0,0, //clubs 0,0,0,0,0,0,0,0,0,0,0,0,0, //diamonds 0,1,1,0,1,0,0,0,0,0,0,0,0, //hearts 0,0,0,1,0,0,0,0,1,0,0,0,0];//spades </code></pre> <p>A 1 represents a card in that value slot. The above hand has a 2-clubs, no diamonds, a 3-hearts, 4-hearts, and 6-hearts, a 5-spades and a 10-spades. Now I look at it to find a straight.</p> <pre><code>var suits = []; //array to hold representations of each suit for (var i=0; i&lt;4; i++) { var index = i*13; // commenting this line as I removed the rest of its use to simplifyy example //var hasAce = (rawHand[i+13]); //get a "suited" slice of the rawHand, convert it to a string representation //of a binary number, then parse the result as an integer and assign it to //an element of the "suits" array suits[i] = parseInt(rawHand.slice(index,index+13).join(""),2); } // OR the suits var result = suits[0] | suits[1] | suits[2] | suits[3]; // Store the result in a string for later iteration to determine // whether straight exists and return the top value of that straight // if it exists; we will need to determine if there is an ace in the hand // for purposes of reporting a "low ace" straight (i.e., a "wheel"), // but that is left out in this example var resultString = result.toString(2); //Show the result for the purposes of this example alert("Result: " + resultString); </code></pre> <p>The trick here is to OR the various suits so there is just one 2-to-Ace representation. Am I wrong in thinking there must be a simpler way to do this?</p>
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    1. COLooks remarkably simple to me, how much simpler are you thinking it could be? For the purpose of detecting the wheel, I'd just insert a copy of the last element of each suit-array at the front of the array before or-ing them together. Then you can just search the string result for '11111'.
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    2. COit could be faster with using a reverse while/for loop `for(var i=3;i--;){...}`
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      1. USvol7ron
    3. CO@Larry Lustig: It's the parseInt line I was thinking could be simpler. You know, as in "maybe if I was better at this I wouldn't have to convert to an array => string => int and then back to a string" ...
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